# Notch Digital Filter

I am given a notch digital filter with the $$z$$-transform being: $$W(z)=MF(z)F(z^{*})^{*}=M\frac{z-q}{z-p}\frac{z-q^{*}}{z-p^{*}}$$ where $$M$$ is the normalisation factor, $$q=e^{-i2\pi\frac{f_0}{f_s}}$$, $$p=(1-\epsilon)q$$ and $$0< \epsilon \ll 1$$. It is easy to see the zeros of this system are on the unit circle while the poles are inside the unit circle.

But how do we determine its stability? I think we must know whether this system is causal or anti-causal. I am very new to this subject so please be clear.

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• Welcome to SE.SP! It's more usual to have stable digital filters with poles inside the unit circle. Where did you have your formulation from? See this page for a python example that gives the magnitude of the poles as array([0.96905238, 0.96905238])
– Peter K.
Mar 18 at 18:35
• Hopefully, that isn't your home work. Mar 19 at 11:02
• With $\epsilon$ fixed in your question, we know that this IIR filter is stable and causal. Mar 19 at 22:50
• A FIR filter is always stable whether it is causal, non-causal or even anti-cause. In the IIR case I don't know, whether there is a relationship between a filter being stable/unstable and its causality ... (but I doubt it). Mar 20 at 8:10

You probably know that a LTI filter is stable if and only if the poles of its transfer function are inside the unit circle.

So I guess you ask for a direct way of showing it in the case of your filter.

The transfer function of a filter is the $$z$$-transform of its impulse response. To show the filter is stable or unstable we have to show that its impulse response does or doesn't converge to zero over time.

Therefore we have to calculate the impulse response, given the transfer function of your filter.

To make things easier we first rewrite $$W(z)$$ as

$$W(z)=M\frac{(z-q)(z-q^*)}{(z-p)(z-p^*)}=M\frac{(1-qz^{-1})(1-q^*z^{-1})}{(1-pz^{-1})(1-p^*z^{-1})}$$

and knowing that convolution in the time domain leads to multiplication in the $$z$$-domain, we will just look at the filter with the transfer function

$$H(z) = \frac{1}{(1-pz^{-1})(1-p^{*}z^{-1})}$$

setting

$$F(z) = \frac{1}{1-pz^{-1}}\text{, }G(z) = \frac{1}{1-p^{*}z^{-1}}$$

we get

$$H(z)=F(z)G(z)\text{.}$$

We use the geometric series and write

$$\sum_{l=0}^{\infty}p^lz^{-l} = F(z)\text{ and } \sum_{l=0}^{\infty}(p^*)^lz^{-l} = G(z)\quad (|z|<|p|)\text{.}$$

Therefore the impluse response for the filter with transfer function $$F$$ is $$f_n = p^n\quad (n\in\mathbb{N})$$ and for the filter with transfer function $$G$$ it is $$g_n = (p^*)^n\quad (n\in\mathbb{N})$$.

Now we calculate the impulse response of the filter with transfer function $$H$$ by $$f\circledast g$$:

$$(f\circledast g)_n = \sum_{l=0}^{n-1}f_lg_{n-l} = \sum_{l=0}^{n-1}p^l(p^*)^{n-l}$$

Using polar representation for $$p = |p|e^{i\varphi}$$ and a geometric series we can deduce

$$(f\circledast g)_n = \sum_{l=0}^{n-1}|p|^le^{il\varphi}|p|^{n-l}e^{-i(n-l)\varphi} =|p|^ne^{-in\varphi}\sum_{l=0}^{n-1}e^{2il\varphi} =\left\lbrace\begin{array}{lr} |p|^nn\text{,} & \text{for } \varphi = 0 \\ |p|^ne^{-in\varphi}\frac{1-e^{2in\varphi}}{1-e^{2i\varphi}} & \text{for } 0<\varphi <\pi\\ (-1)^n|p|^nn\text{,} & \text{for } \varphi = \pi \end{array}\right\rbrace$$

which converges to $$0$$ for $$t\to\infty$$ if and only if $$|p| < 1$$. So for $$|p| < 1$$ the filter is stable and for $$|p| \ge 1$$ the filter is unstable.

Now you can find the impulse response $$(l_n)_{n=0}^\infty$$ for the FIR filter $$\mathcal{L}$$ with transfer function $$L(z) = M(1-qz^{1})(1-q^*z^{-1})$$ (which is not so hard) and convolve it with $$f\circledast g$$ and deduce some inequalities to show $$(l\circledast (f\circledast g))$$ converges to $$0$$ over time if and only if $$|p| < 1$$.

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typetetris is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• I am not a native speaker, so sorry for any english mistakes. Mar 19 at 11:03
• For the Z-transform you can trade-off stability for causality. It depends on where you choose the ROC to be. Since the OP specifically mentioned causality, your answer should address this. Mar 20 at 15:58
• True. I alter my answer when I am off work. Mar 21 at 8:46